# Upper and lower bounds on reciprocals of restricted prime products

Let $k$ be fixed and consider the sum $$F(k,n)=\sum_{p_1<p_2<\cdots<p_k\leq n~:~p_1 p_2\cdots p_k\leq n} \frac{1}{p_1 p_2 \cdots p_k}.$$ Are tight upper and lower bounds on $F(k,n)$ known as $n\rightarrow \infty$? One approximation to the sum might be $$\left( \sum_{p\leq n^{1/k}} \frac{1}{p}\right)^k\approx \left[\log \left (\frac{1}{k} \log n\right)\right]^k=\left[\log \log n -\log k\right]^k$$but I am unsure how good this approximation is.

Now, let $k$ vary very slowly with $n$, say $k=c \log \log n.$ Do the answer(s) (if known) change?

## 1 Answer

Let $\pi_k(x)$ be the number of integers $\leq x$ with exactly $k$ prime factors. In the range $k<e\log\log x$, an asymptotic formula for $\pi_k(x)$ was given by Sathe (J. Indian Math. Soc. (N.S.) 17, (1953), 63–82). Selberg (J. Indian Math. Soc. (N.S.) 18, (1954), 83–87) gave a much simpler proof. Sifting out non-squarefree values and applying partial summation you obtain an asymptotic formula for $F(k, n)$ in the same range.

For small values of $k$, your argument actually gives an asymptotic formula. If $G(k, n)$ denotes the sum without the restriction that the $p_i$ are different and increasing, we have $$\left(\sum_{p\leq n^{1/k}}\frac{1}{p}\right)^k\leq G(k, n)\leq\left(\sum_{p\leq n}\frac{1}{p}\right)^k,$$ thus the known bound for $\sum_{p\leq x}\frac{1}{p}$ give an asymptotic formula for $G(k, n)$, provided that $(\log\log n-\log k)^k\sim(\log\log n)^k$, that is, $k=o(\frac{\log\log n}{\log\log\log n})$. Passing from $G$ to $F$ takes some work, but is elementary.

• Thanks. I don't understand your two inequalities, is there meant to be exponents on either expression? – kodlu Dec 16 '15 at 20:34
• Yes, the exponent $^{1/k}$ in the left sum was missing. – Jan-Christoph Schlage-Puchta Dec 17 '15 at 21:32
• Thanks. I can't access the Sathe or Selberg papers. Could you suggest a number theory text or a more accessible article which covers the derivation you refer to as "elementary"? – kodlu Dec 17 '15 at 22:10
• @Jan-Cristoph Schlage-Puchta, In particular, I can see the left sum is asymptotically $\log \log(n^{1/k})=-\log(k)+\log \log n$ (ignoring the Massel-Merteens constant) which becomes $\log\log n -\log\log\log n$ if $k=\log\log n.$ Is this correct? It's the upper bound that is more involved, I think. – kodlu Dec 17 '15 at 22:26
• Now it should be correct. You essentially get $(\log\log n-\log k)^k\leq G(k,n)\leq (\log\log n)^k$. To pass from $G$ to $F$ you have to delete terms with repeating primes, and divide by $k!$ to take care of the ordering. For small $k$ you can bound the contribution of non-squarefree integers by $G(n,k-2)\binom{k}{2}\sum_p\frac{1}{p^2}$. (Pick a prime $p$ which occurs twice, pick the places where it occurs.) For $k$ much smaller than $\sqrt{\log\log n}$ this gives an asymptotic formula for $F(n,k)$. As $k$ gets larger, things get messier. – Jan-Christoph Schlage-Puchta Dec 17 '15 at 22:39